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3 years ago

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How much heat is required to increase the temperature of 10.0 grams of water 6.0oC? (The specific heat of water is 4.18 J/g x oC

)

1 answer:

Paraphin [41]3 years ago

5 0

250.8 J

Approximately 250 J (A on Plato)

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At equilibrium, no chemical reactions take place. true or false

arsen [322]

False. At equilibrium, the rate of forward reaction is equal to the rate of backward reaction. The net concentration of both products and reactants won't change, but the reactions still take place.

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2 years ago

Rocks that are melted by heat and then cooled are classified as: igneous sedimentary metamorphic.

Gnom [1K]

The answer is igneous

8 0

3 years ago

Read 2 more answers

According to the following reaction, how many grams of nitrogen monoxide will be formed upon the complete reaction of 31.0 grams

Mumz [18]

23.227 grams of NO is formed upon the complete reaction of 31.0 grams of oxygen gas with excess ammonia.

Explanation:

The balanced chemical equation is:

4NH3 + 5O2 ⇒ 4NO + 6H20

Number of moles will be calculated.

31 grams of oxygen is given

atomic weight of oxygen gas i.e O2 = 32gm/mole

The molar mass of oxygen gas is 32 grams/mole which is equal to one mole of the molecule.

Number of moles (n) = $\frac{mass}{atomic mass of one mole of the substance}$

n= $\frac{31}{32}$

n = 0.968 moles of oxygen are given for the reaction to occur.

From the balanced chemical equation it can be seen

5 moles of oxygen yielded 4 moles of NO

So, 0.968 moles of oxygen yield x moles of NO

$\frac{4}{5}$ = $\frac{x}{0.968 }$

4 × 0.968 =5x

x = 0.774 moles of NO will be formed.

To calculate the mass of NO

mass= atomic mass x number of moles

atomic mass of NO is 30.01 grams/mole

putting the values in the formula:

mass = 30.01 x 0.774

= 23.227 grams of NO is formed.

8 0

3 years ago

Convert<br> 12 x 1025 kg/mL to fg/cm3

Anna35 [415]

Answer:

$1.2x10^{38}\frac{fg}{cm^3}$

Explanation:

Hello there!

In this case, for such unit conversion we need to realize that 1 kg is equal to 1000 g, 1 g is equal to 1x10⁹ and 1 mL equals 1 cm³, therefore we apply:

$12x10^{25}kg*\frac{1000g}{1kg} *\frac{1x10^9fg}{1g} *\frac{1mL}{1cm^3} \\\\1.2x10^{38}\frac{fg}{cm^3}$

Best regards!

5 0

3 years ago

A three level system has energies ε0=0.00J, ε1=5.00x10-21J, and ε2=10.0x10-21J. a. Using the Boltzmann Distribution Law, calcuat

choli [55]

Answer:

A three level system has energies ε0=0.00J, ε1=5.00x10-21J, and ε2=10.0x10-21J. a. Using the Boltzmann Distribution Law, calcuate the molar internal energy U at T=10.0°C. Hint: This means you have to calculate three probabilites at T=10K and use them to calculate the average energy per particle and then calculate the energy of a mole of particles.b. Again using the Bolztmann Distribution Law and the same approach as in part A, calculate the molar internal enerrgy U at T=1000°C c. Using you results from parts A and B, calculate the change in the molar internal energy ΔU when the temperature changes from T=10°C to T=1000°C.d. Using the Bolzmann Distribution Law probabilities calculated in part A, calculate the molar entropy at T=10°C.e. Calculate the change in molar entropy when the temperature of this 3 level system changes from T=10°C to T = 1000°C

A). 960.9J/mol

B). 2447.31J/mol

C). 1486.41J/mol

D). 577.706JK-1mol-1

E). 576.233 JK-1mol-1

F). 1.473JK-1mol-1

Explanation:

partition function q equals

(a) q = 1+e(−ε1/kT) + γe(−2ε2/kT) .

As such substituting the values for the energies at the different states we have where putγ= 1

q = 1 + 0.2779+7.726 = 1.3552

From where we have populations, or probabilities, P1, P2 , P3

P1 = 1/q= 1/1.3552= 0.7378

P2 = e−ε1/kT/q = 0.2051

And P3 = e−2ε2/kT/q = 0.05701

The Average energy per particle can be found as

[ε0] = 0P0 + ε0Pi + ε1 P2

= 0 + 5×10-21 ×0.2051 + 10×10-21× 0.05701 = 1.5956×10-21J

1.5956×10-21J×6.02×1023 particles= 960.9J/mol

B). at 1000 C we have also

q = 1+e(−ε1/kT) + e(−2ε2/kT)

From where q = 1+0.7523+0.5659 = 2.31825

From where P1 = 0.43136, P2 = 0.3245 and P3 = 0.244

And the average energy is

[ε0] = 0P0 + ε0Pi + ε1 P2

= 4.0638×10-21J hence U = 4.0638×10-21J × 6.02×1023 particles = 2447.31J/mol

C). Change in molar internal energy between 10 and 1000C =

2447.31J/mol-960.9J/mol = 1486.41J/mol

D). at 10C we have

S/n = R×lnq + ∆U/(n×T)

∆U/n = 960.9J/mol hence

S/n = (8.314JK-1mol-1) ln(1030) + 960.9J/(283 mol K)

= S/n = 577.706JK-1mol-1

E). At 1000C we have

S/n = (8.314JK-1mol-1) ln(1030) + 2447.31J/(1273 mol K) = 576.233 JK-1mol-1

F). Difference = 577.706JK-1mol-1 - 576.233 JK-1mol-1 = 1.473JK-1mol-1

5 0

3 years ago